Antiderivatives

An antiderivative is useful when you know the rate of change, and you want to find a point from that rate of change

A function F is said to be an antiderivative of f if F'(x) = f(x)

Notation

The collection of all antiderivatives of a function f is referred to as the indefinite integral of f with respect to $x$, and is denoted by:

\int f(x) dx

Find the antiderivative of the function y = x^2

We know that to find the derivative of the above function, you'd multiply by the exponent (2), and subtract 1 from the exponent.
To perform this operation in reverse:
1. Add 1 to the exponent
2. Multiply by \dfrac{1}{n + 1}
This gives us an antiderivative of \dfrac{1}{3}x^3
To check our work, work backwards.
The derivative of \dfrac{1}{3}x^3 is \dfrac{1}{3} (3x^2)
= \dfrac{3}{3} x^2

Differentiation Formula	Integration Formula
`\dfrac{d}{dx} x^n = nx^{x-1}`	`\int x^n dx = \dfrac{1}{n+1}x^{n+1}+ C` for `n \ne -1`
`\dfrac{d}{dx} kx = k`	`\int k \space dx = kx + C`
`\dfrac{d}{dx} \ln \\|x\\| = \dfrac{1}{x}`	`\int \dfrac{1}{x}dx = \ln \\|x\\| + C`
`\dfrac{d}{dx} e^x = e^x`	`\int e^x dx = e^x + C`
`\dfrac{d}{dx} a^x = (\ln{a}) a^x`	`\int a^xdx = \ln \\|x\\| + C`
`\dfrac{d}{dx} \sin x = \cos x`	`\int \cos(x) dx = \sin (x) + C`
`\dfrac{d}{dx} \cos x = -\sin x`	`\int \sin(x)dx = \sin x + C`
`\dfrac{d}{dx} \tan{x} = \sec^2 x`	`\int \sec^2(x)dx = \tan(x) + C`
`\dfrac{d}{dx} \sec x = \sec x \tan x`	`\int sec^2(x) dx = \sec(x) + C`
`\dfrac{d}{dx} \sin^{-1} x = \dfrac{1}{\sqrt{1-x^2}}`	`\int \sec(x) \tan(x) dx = \sec x + C`
`\dfrac{d}{dx} \tan^{-1} x = \dfrac{1}{1+x^2}`	`\int \dfrac{1}{\sqrt{1+x^2}}dx = \tan^{-1}x + C`
`\dfrac{d}{dx} k f(x) = k f'(x)`	`\int k*f(x)dx = k\int f(x)dx`
`\dfrac{d}{dx} f(x) \pm g(x) = f'(x) \pm g'(x)`	`\int (f(x) \pm g(x))dx = \int f(x) dx \pm \int g(x) dx`